Several different methods have been used to time synchronize the oscillators or clocks in two or more pieces of equipment that are in communication with one another. For example, a reference transmitter has been used to time synchronize the oscillators or clocks in two or more remote units (RUs) in some existing multilateration systems.
One example of a time synchronization protocol is IEEE 1588, which defines a messaging protocol for synchronizing real-time clocks in the nodes of a distributed networked system. IEEE 1588 describes a hierarchical master-slave clock architecture for time synchronization and clock distribution. In the IEEE 1588 architecture, a time distribution system consists of one or more network segments with one or more clocks in each segment, and each clock is defined as a master clock, a slave clock or a boundary clock.
The IEEE 1588 Precision Time Protocol (PTP) provides a standard message exchange approach that allows a slave clock to be synchronized to a master clock, as shown in FIG. 1. A master clock is selected for each of the network segments in the system and is used as the time synchronization source. A boundary clock is a clock that has multiple network connections and can accurately bridge the time synchronization from one network segment to another network segment. A boundary clock acts as a slave clock in one network segment and a master clock in another network segment. Other clocks in the distributed networked system are slave clocks that are the destination of the time synchronization reference.
IEEE 1588 requires at least one message being sent from the master node to the slave node and one message being sent from the slave to the master for each time synchronization process, as shown in FIG. 1. At the end of one time synchronization process, the slave node has in its possession four pieces of time information, namely time-stamp data for T1, T2′, T3′, and T4, while the master node has only partial time information, namely time-stamp data for T1 and T4.
Assuming throughout the time synchronization process the distance or propagation delay between the two nodes remains symmetric (i.e. T2−T1=T4−T3) and the clock offset remains constant (i.e. bs does not vary with time), at the end of the time synchronization process the Slave node is able to calculate its clock offset bs and the range R to the master node, based on following equations
                                          R            ≡                                                            T                  ⁢                                                                          ⁢                  2                                -                                  T                  ⁢                                                                          ⁢                  1                                            C                                =                                                    T                ⁢                                                                  ⁢                4                            -                              T                ⁢                                                                  ⁢                3                                      C                          ⁢                                  ⁢                                                                              ⇒                  R                                =                                ⁢                                                                                                    [                                                                              (                                                                                          T                                ⁢                                                                                                                                  ⁢                                2                                                            +                              bs                                                        )                                                    -                                                      T                            ⁢                                                                                                                  ⁢                            1                                                                          ]                                            +                                              [                                                                              T                            ⁢                                                                                                                  ⁢                            4                                                    -                                                      (                                                                                          T                                ⁢                                                                                                                                  ⁢                                3                                                            =                              bs                                                        )                                                                          ]                                                                                    2                      ⁢                      C                                                        =                                    ⁢                                                                                    (                                                                              T                            ⁢                                                                                                                  ⁢                                                          2                              ′                                                                                -                                                      T                            ⁢                                                                                                                  ⁢                            1                                                                          )                                            +                                              (                                                                              T                            ⁢                                                                                                                  ⁢                            4                                                    -                                                      T                            ⁢                                                                                                                  ⁢                                                          3                              ′                                                                                                      )                                                                                    2                      ⁢                      C                                                                                                                                              bs                =                                ⁢                                                                                                    [                                                                              (                                                                                          T                                ⁢                                                                                                                                  ⁢                                2                                                            +                              bs                                                        )                                                    -                                                      T                            ⁢                                                                                                                  ⁢                            1                                                                          ]                                            -                                              [                                                                              T                            ⁢                                                                                                                  ⁢                            4                                                    -                                                      (                                                                                          T                                ⁢                                                                                                                                  ⁢                                3                                                            +                              bs                                                        )                                                                          ]                                                              2                                    =                                    ⁢                                                                                    (                                                                              T                            ⁢                                                                                                                  ⁢                                                          2                              ′                                                                                -                                                      T                            ⁢                                                                                                                  ⁢                            1                                                                          )                                            +                                              (                                                                              T                            ⁢                                                                                                                  ⁢                            4                                                    -                                                      T                            ⁢                                                                                                                  ⁢                                                          3                              ′                                                                                                      )                                                                                    2                      ⁢                      C                                                                                                                              (        1        )            where C is the speed of light.
Most systems for determining own position require the transmission and receipt of dedicated signals for position determination. With the density of transmitted signals being transmitted, what is needed is a system and method that enables a mobile node to determine its own position without transmitting any signals by passively listening to wireless time synchronization communications, such as IEEE 1588, between a plurality of wireless transceivers.